I was recently working on a quadcopter so I bought a pair of coreless DC motor. My intention was to use an Arduino (DIP chip) to control the speed of the DC motor (rated 3.7 V) through a transistor.
But I have noticed that whenever I connect the motor to the battery (li-po 3.7 V) the voltage drops to about 2.8 V. I have tried a 5 V boost converter but the voltage dropped to 2.8 V again.
Can anyone please tell me the reason.
This is known as the internal resistance of the battery. As soon as you draw current, the voltage at the battery's leads drops. Nevertheless, only 2.8 V sounds really little, maybe your battery is just drained.
EDIT: Come to think about it, a fully charged 15C battery can't easily be stressed to the point where it only outputs 2.8V. Especially not by such a puny motor. Are you sure the battery isn't just dead?
It sounds like the motor is trying to draw more current than the battery can supply. The battery is like a squirt gun, and the motor needs a fire hose.
A Li-PO battery is sold and is stored at a partial 3.7V charge. It is 4.2V after a proper charge when it has full power. A proper charge is when it continues charging at 4.2V until its charging current drops to 1/10th or less.
My Li-PO batteries slowly drive my motors and not for long when they have a 3.7V storage charge.
A Li-PO battery must never be stored fully charged. If yours have been fully charged for longer than a few months then they are ruined and will produce low power.
A Li-PO also must never be discharged below about 3.0V or again it will be ruined.
add a circuit like this between the Arduino chip and the motor. If I'm correct, the voltage will decrease due to the resistor, but the current should be the same as the battery, and will balance out. the concept is that if you put capacitors in parallel they all charge and create more current. You may need to experiment with the resistor a bit. also tell me if it doesn't work, I made this circuit quickly.
